What is the irrational cupe root ³√40 rewritten as a product of an integer and another irrational cube root? (1 point)
O 2 ³√5
O - 2 ³√5
O 8 ³√5
O 2 ³√10
The answer is O 2 ³√10.
To rewrite the irrational cube root ³√40 as a product of an integer and another irrational cube root, we first need to find a perfect cube that is a factor of 40. The largest perfect cube that is a factor of 40 is 8, which is 2^3.
Therefore, we can rewrite ³√40 as ³√(8 * 5).
This can further be simplified as 2 * ³√5.
So, the answer is O 2 ³√5.
To rewrite the irrational cube root ³√40 as a product of an integer and another irrational cube root, we need to find a perfect cube that can be factored out from 40.
Let's analyze the factors of 40:
40 = 2 × 2 × 2 × 5
Since there are three 2's in the factorization, we can take one of them outside the cube root as 2³√(2 × 2 × 2). This simplifies to 2³√8.
The remaining factors inside the cube root are 2 and 5. Therefore, ³√40 can be rewritten as 2³√(8 × 5).
So, the correct answer is: O 2 ³√(8 × 5)